Rees Modules Artin Rees and Hilbert Samuel Theory — Examples

In $k[x,y]$, the degree-$n$ monomials are exactly $x^{n-i}{y}^i$ for $0\le i\le n$, so the degree-$n$ piece has dimension $n+1$. Therefore ${\mathrm{HS}}_{k[x,y]}(t)={\sum }_{n\ge 0}(n+1)t^n=\frac{1}{(1-t{)}^2}\mathrm{.}$

In the quotient by $(y^2)$, every monomial containing $y^2$ vanishes. So for $n\ge 1$ the degree-$n$ piece is spanned by $x^n$ and $x^{n-1}y$, and these two classes are linearly independent. Hence ${\mathrm{HS}}_A(t)=1+{\sum }_{n\ge 1}2t^n=\frac{1+t}{1-t}\mathrm{.}$

The eventual coefficient sequence of ${\mathrm{HS}}_A(t)$ is constant equal to $2$, so its Hilbert polynomial is $2$.

Thus both the polynomial ring and this homogeneous quotient realize the expected Hilbert series and Hilbert polynomial.

In $R_1$, the classes of $x$ and $y$ in $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ generate every graded piece: the images of the degree-$n$ monomials $x^{n-i}{y}^i$ form a basis of ${\mathfrak{m}}^n\mathrm{/}{\mathfrak{m}}^{n+1}$. Therefore the map $k[X,Y]\to {\mathrm{gr}}_{\mathfrak{m}}(R_1)$ sending $X,Y$ to the initial classes of $x,y$ is a graded isomorphism.

In $R_2$, the relation $y^2-x^3$ lies in ${\mathfrak{m}}^2$ and its lowest-degree term is $y^2$. Hence the only initial relation in degree $2$ is $Y^2=0$. As in the remaining monomials $X^n$ and $X^{n-1}Y$ survive and span the graded pieces, so ${\mathrm{gr}}_{\mathfrak{m}}(R_2)\cong k[X,Y]\mathrm{/}(Y^2)\mathrm{.}$

These explicit computations exhibit the regular local and cusp cases.

Here $I^{n}M=(x^n)$ and $N=(x^m)$. If $n\ge m$, then $(x^n)\subseteq (x^m)$, so $I^{n}M\cap N=(x^n)\mathrm{.}$

Also $I^{m}M\cap N=(x^m)$, and therefore for $n\ge m$, $I^{n-m}(I^{m}M\cap N)=(x^{n-m})(x^m)=(x^n)=I^{n}M\cap N\mathrm{.}$

So $c=m$ works, exhibiting an explicit Artin-Rees bound compatible with the abstract existence statement [L1].

View $R$ as a finite module over itself. Since $(R,\mathfrak{m})$ is local, its maximal ideal lies in the Jacobson radical. Therefore [L1] applies with $M=R$ and gives ${\bigcap }_{n\ge 0}{\mathfrak{m}}^n=0.$

This is the promised local-domain instance of Krull intersection.

By [L1], the Hilbert-Samuel function is ${\chi }_{(\pi ),V}(n)={\mathrm{\ell }}_V(V\mathrm{/}({\pi }^{n+1}))=n+1$ for every $n\ge 0$.

Therefore the eventual polynomial is already exactly $P(n)=n+1$. Its degree is $1$ and its leading coefficient is $1$, so [L2] gives $e_{(\pi )}(V)=1!\cdot 1=1.$

This computes both the Hilbert-Samuel polynomial and the multiplicity of a DVR.

The initial form of $y^2-x^3$ has degree $2$, namely $Y^2$, so ${\mathrm{gr}}_{\mathfrak{m}}(R)\cong k[X,Y]\mathrm{/}(Y^2)\mathrm{.}$ Thus the degree-$0$ piece has dimension $1$, and every degree-$n\ge 1$ piece has basis $X^n,X^{n-1}Y$.

Therefore ${\mathrm{\ell }}_R({\mathfrak{m}}^0\mathrm{/}\mathfrak{m})=1$ and ${\mathrm{\ell }}_R({\mathfrak{m}}^n\mathrm{/}{\mathfrak{m}}^{n+1})=2(n\ge 1)\mathrm{.}$ Summing these lengths as in [L1] gives ${\chi }_{\mathfrak{m},R}(n)=2n+1$ for $n\ge 1$.

The eventual polynomial is $2n+1$, so its degree is $1$ and the leading coefficient is $2$. Hence [L2] gives $e_{\mathfrak{m}}(R)=2$.

Thus the multiplicity of the cusp is read directly from its tangent-cone graded ring.

In a zero-dimensional local ring every prime ideal is maximal, hence the only prime is $\mathfrak{m}$. Its nilradical is therefore $\mathfrak{m}$, and [L2] gives ${\mathfrak{m}}^N=0$ for some $N\ge 1$. Hence for every $n\ge N-1$, $M\mathrm{/}{\mathfrak{m}}^{n+1}M=M$.

The finite module $M$ has finite length over the zero-dimensional Noetherian local ring: the filtration by powers of the nilpotent maximal ideal has finitely many quotients, each a finite-dimensional vector space over $R\mathrm{/}\mathfrak{m}$. Thus ${\mathrm{\ell }}_R(M)$ is defined by [L1]. Step 1.1 shows that the Hilbert-Samuel function is eventually the constant polynomial ${\mathrm{\ell }}_R(M)$, uniquely so because two polynomials agreeing at all large integers are equal.

This is exactly the zero-dimensional case.